Notes - math

Lecture notes for Topology MMA100
J A S, S-11
The fundamental group
A path α from the point x to the point y in a space X is a map α : I → X, such that α(0) = x and
α(1) = y. We find it convenient to write this symbolically as x → y, though α is of course not a map
from the point x to the point y.
If x → y, then α−1 is the path y → x defined by α−1 (t) = α(1 − t).
The constant path cx at the point x in X is defined by cx (t) = x for all t ∈ I. It is a path x → x.
Paths x → y → z can be composed to a path x → z, by defining
if 0 ≤ t ≤ 1/2
α.β(t) =
β(2t − 1) if 1/2 ≤ t ≤ 1.
The continuity of α.β follows from the Gluing lemma in Armstrong’s book (or can easily be checked by
This composition is not associative in general; (α.β).γ means doing γ half the time, whereas α.(β.γ)
means doing γ a quarter of the time.
If f : X → Y is a map and x → y is a path in X, then f ◦ α is a path f (x) → f (y) in Y .
We note that it is immediate from the definition of the composition that f ◦ (α.β) = (f ◦ α).(f ◦ β).
Homotopy classes of paths
α, β
Two paths x → y in X are (path-)homotopic, α ' β, if there is a map F : I × I → X, such that, for all
t and s,
F (t, 0) = α(t), F (t, 1) = β(t), F (0, s) = x, and F (1, s) = y.
The map F is then called a (path-)homotopy from α to β, written F : α ' β. Note that if s is fixed the
map Fs (t) = F (t, s) is a path x → y and, with this notation, α = F0 , while β = F1 .
Theorem 1.1 The relation ' is an equivalence relation on the set of paths x → y.
Proof. Suppose F : α ' β. Define F −1 (t, s) = F (t, 1 − s). Then F −1 : β ' α. This shows symmetry.
Suppose F : α ' β and G : β ' γ. Define
F (t, 2s)
F.G(t, s) =
G(t, 2s − 1)
0 ≤ s ≤ 1/2
1/2 ≤ s ≤ 1.
Note that F.G is defined in two different manners on two closed subsets of I 2 = I × I, which agree on
the intersection of the two closed subsets (where s = 1/2). Thus F.G is indeed a function I 2 → X. It is
evidently continuous when restricted to each of the two closed sets and hence continuous on all of I 2 by
an application of the Gluing lemma. Now F.G : α ' γ. This shows transitivity.
Finally, let F (t, s) = α(t). Then F : α ' α. This shows reflexivity.
We denote the equivalence class of x → y by hαi and sometimes write x → y to emphasize that any
representative of hαi is a path x → y.
A subspace C of Rn is convex if all points on the ray (1 − t)p + tq, 0 ≤ t ≤ 1, are in C if p and q are. We
note that in a such a space all paths p → q in C are homotopic. Indeed, let α and β be two such paths.
Define F (t, s) = (1 − s)α(t) + sβ(t). The construction is possible by the linear structure of Rn and the
image of F is in C by convexity. Then F : α ' β.
Furthermore, if f : X → Y is a map and F : α ' β is a homotopy between paths in X, then f ◦ F is a
homotopy f ◦ α ' f ◦ β. This allows us to make the definition f∗ (hαi) = hf ◦ αi.
β,β 0
Theorem 1.2 Suppose x → y → z are paths in a space X, that α ' α0 and β ' β 0 . Then α.β ' α0 .β 0 .
Proof. Suppose F : α ' α0 and G : β ' β 0 . Define F∗ G : I 2 → X by
F (2t, s)
if 0 ≤ t ≤ 1/2
F∗ G(t, s) =
G(2t − 1, s) if 1/2 ≤ t ≤ 1.
Continuity of F∗ G follows as above by appealing to the Gluing lemma. Now F∗ G : α.α0 ' β.β 0 .
Using this we can safely define composition of x → y and y → z as hαi.hβi = hα.βi.
Theorem 1.3 Suppose that x → y → z → w are paths in X. Then
1. α.(β.γ) ' (α.β).γ, so that hαi.(hβi.hγi) = (hαi.hβi).hγi
2. cx .α ' α ' α.cy , so that hcx i.hαi = hαi = hαi.hcy i,
3. cx ' α.α−1 , so that hcx i = hαi.hα−1 i.
Proof. Define a map f : [0, 3] → X by
 α(t)
β(t − 1)
f (t) =
γ(t − 2)
2 ≤ t ≤ 3.
Continuity of f follows as usual. Define paths 0 →
1, 1 →
2 and 2 →
3 in the convex space [0, 3], by
1 (t) = t, 2 (t) = t + 1 and 1 (t) = t + 2. Then f ◦ 1 = α, f ◦ 2 = β and f ◦ 3 = γ. Convexity of [0, 3]
gives 1 .(2 .3 ) ' (1 .2 ).3 since both are paths 0 → 3. By applying f to this we get α.(β.γ) ' (α.β).γ.
This proves 1.
To prove 2 and 3, define f : [0, 1] → X by f (t) = α(t) (to avoid confusion). Then, as above, f ◦ 1 = α,
f ◦ −1
while f ◦ c0 = cx and f ◦ c1 = cy .
1 =α
By convexity of [0, 1] we have c0 .1 ' 1 ' 1 .c1 , since all three are paths 0 → 1. Applying f to this gives
2. Similarly, c0 ' 1 .−1
1 , since both are paths 0 → 0 in the convex space [0, 1]. Applying f to this gives
The fundamental group(s)
Let p be a point in a space X. We denote by π1 (X, p) the set of homotopy classes of paths p → p in X.
A path p → p is called a loop in X based at p.
We can define a composition rule on this set by defining hαi.hβi = hα.βi. From a theorem above we
know that this is an associative composition and that hcp i.hαi = hαi = hαi.hcp i, so that the class of the
constant loop at p is a unit element for the composition rule. We also know that hαi.hα−1 i = hcp i, so
that hα−1 i is an inverse of hαi with respect to the composition, i.e. hα−1 i = hαi−1 .
With this we have given the set π1 (X, p) the structure of a group. It is called the fundamental group of
X based at p (the base point).
We check how this group depends on the choice of base point.
Theorem 1.4 Let p → q be a path in X. Then the function ω∗ : π1 (X, p) → π1 (X, q) defined by
ω∗ (hαi) = hω −1 i.hαi.hωi is an isomorphism of groups.
Proof. We need to check that ω∗ (hαi.hβi) = ω∗ (hαi).ω∗ (hβi) and that ω∗ hcp i = hcq i, so that ω∗ maps
the unit element to the unit element.
ω∗ (hαi.hβi)
= hω −1 i.hαi.hβi.hωi =
= hω −1 i.hαi.hcp i.hβi.hωi =
= hω −1 i.hαi.hωi.hω −1 i.hβi.hωi =
= ω∗ (hαi).ω∗ (hβi).
ω∗ (hcp i) = hω −1 i.hcp i.hωi = hω −1 i.hωi = hcq i.
Finally, the homomorphism (ω −1 )∗ : π1 (X, q) → π1 (X, p) is inverse to ω∗ so that this function is an
isomorphism of groups.
From this result it follows that if a space is path-connected then any two fundamental groups of that space
are isomorphic.
Next we consider the effect of a map f : X → Y on fundamental groups.
Theorem 1.5 If f : X → Y is a map, then the function f∗ : π1 (X, p) → π1 (Y, f (p)) defined by f∗ (hαi) =
hf ◦ αi is a group homomorphism.
Proof. We have
f∗ (hαi.hβi)
f∗ (hα.βi) =
hf ◦ (α.β)i = h(f ◦ α).(f ◦ β)i =
hf ◦ αi.hf ◦ βi = f∗ (hαi).f∗ (hβi),
f∗ (hcp i)
= hf ◦ cp i = hcf (p) i.
It follows from this result that if f is a homeomorphism then f∗ is an isomorphism with inverse given by
(f −1 )∗ . Thus homeomorphic spaces have isomorphic fundamental groups.
A space X is simply connected if it is path-connected and π1 (X, x) is a trivial group for some (hence any)
x ∈ X.
The homotopy relation on maps and the fundamental group
In the last section we saw that spaces of the same homeomorphism type (i.e homeomorphic spaces) have
isomorphic fundamental groups. In this section we introduce a cruder equivalence relation – being of the
same homotopy type – on the class of all spaces and show that spaces of the same homotopy type have
isomorphic fundamental groups.
Spaces of the same homotopy type can be wildly non-homeomorphic, but the fundamental group cannot
see the difference between such spaces. This can be considered as a draw back if we are interested in
determining if two spaces are homeomorphic or not, but it also has a positive side: it makes the fundamental group easier to calculate. Suppose for example that we would like to compute the fundamental
group of a space X. If X turns out to be of the same homotopy type as a space Y of which we already
know the fundamental group, then the fundamental group of X will be isomorphic to this group.
The homotopy relation
Two maps f, g : X → Y are homotopic, written f ' g, if there is a map F : X × I → Y, such that F0 = f
and F1 = g. Here Fs (x) = F (x, s). We write such a map symbolically as F : f ' g, and say that F is a
homotopy from f to g.
Theorem 2.1 The relation ' is an equivalence relation on the set of maps X → Y .
Proof. Given f : X → Y define F (x, t) = f (x) for all (x, t) ∈ X × I. Then F : f ' f, which shows
Given F : f ' g, define F −1 (x, t) = F (x, 1 − t). Then F −1 : g ' f . This shows symmetry.
Given F : f ' g and G : g ' h define F.G : X × I → Y as
F (x, 2t)
if 0 ≤ t ≤ 1/2
F.G(x, t) =
G(x, 2t − 1) if 1/2 ≤ t ≤ 1.
Thus F.G is defined in two different ways on the closed subset A = X × [0, 1/2] and B = X × [1/2, 1]
of A ∪ B = X × I. We note that the two definitions agree on A ∩ B = X × {1/2} and consequently
do define a function X × I → Y . Furthermore, F.G is evidently continuous on A and B and hence on
A ∪ B = X × I by the Gluing lemma.
We now have F.G : f ' h.
Next we check that the homotopy relation behaves well with respect to composition of maps:
Theorem 2.2 Suppose that f and f 0 are homotopic maps X → Y and that g and g 0 are homotopic maps
Y → Z. Then g ◦ f ' g 0 ◦ f 0 .
Proof. Suppose, more specifically, that F : f ' f 0 and G : g ' g 0 . Then g ◦ F : g ◦ f ' g ◦ f 0 while
G ◦ (f 0 × 1I ) : g ◦ f 0 ' g 0 ◦ f 0 , so g ◦ f ' g 0 ◦ f 0 by transitivity of the homotopy relation on maps X → Z.
A map f : X → Y is said to be a homotopy equivalence if there is a map g : Y → X, called a homotopy
inverse of f, such that
' 1X
f ◦g
' 1Y .
Notice that f is then a homotopy inverse of g, so that g is also a homotopy equivalence. If f is a
homeomorphism and we choose g = f −1 we see that any homeomorphism is a homotopy equivalence.
By the theorem above any map homotopic to a homotopy equivalence is itself a homotopy equivalence.
Also, any two homotopy inverses g and g 0 to f are homotopic; g 0 = g 0 ◦ 1Y ' g 0 ◦ f ◦ g ' 1X ◦ g ' g.
Thus a homotopy equivalence behaves like a homeomorphism if you view it through homotopy glasses.
Two spaces X and Y are homotopy equivalent, or of the same homotopy type, if there is a homotopy
equivalence X → Y . Two spaces of the same homeomorphism type are also of the same homotopy type,
so the latter is a coarser equivalence relation on spaces than the former.
Homotopy and the fundamental group
We next consider how homotopic map relates when we consider their effect on fundamental groups:
Theorem 2.3 Suppose that f, g : X → Y are homotopic and p ∈ X is a choice of base point. Then
there is a path f (p) → g(p) in Y such that the composite ω∗ ◦ f∗ : π1 (X, p) → π1 (Y, f (p)) →∗ π1 (Y, g(p))
equals g∗ : π1 (X, p) → π1 (Y, g(p)).
Recall that the map ω∗ is an isomorphism of groups by a previous theorem. Thus, even though we cannot
say that f∗ and g∗ are the same homomorphism they only differ by an isomorphism.
Proof. Suppose F : f ' g and let p → p be a loop at p in X. Define the map H : I × I → Y to be the
composite of α × 1I : I × I → X × I with F : X × I → Y, so that H(t, s) = F (α(t), s). In I × I define
four paths along the edges as follows:
(1, 0)
(0, 0) →
(0, 1) → (1, 1)
(0, 0) → (0, 1)
(1, 0) → (1, 1)
1 (t) = (t, 0)
2 (t) = (t, 1)
3 (t) = (0, t)
4 (t) = (1, t).
Then, defining ω(t) = F (p, t) we have a path f (p) → g(p) and
H ◦ 1 (t) = F (α(t), 0)
f ◦ α(t)
H ◦ 2 (t) = F (α(t), 1)
g ◦ α(t)
H ◦ 3 (t) = F (α(0), t)
F (p, t) = ω(t)
H ◦ 4 (t) = F (α(1), t)
F (p, t) = ω(t)
Convexity of I × I ⊂ R2 gives that 2 ' −1
3 .(1 .4 ), since both are paths (1, 0) → (1, 1) in I × I.
Applying H to this gives
g ◦ α ' ω −1 .((f ◦ α).ω)
g∗ (hαi)
= hg ◦ αi =
= hω −1 .((f ◦ α).ω)i =
= hω −1 i.hf ◦ αi.hωi =
= ω∗ (hf ◦ αi) = ω∗ ◦ f∗ (hαi).
Since α is arbitrary this shows the claim.
Finally, we show that spaces of the same homotopy type have isomorphic fundamental groups.
Theorem 2.4 Suppose f : X → Y is a homotopy equivalence. Then f∗ : π1 (X, p) → π1 (Y, f (p)) is an
isomorphism for all choices of base points p ∈ X.
Proof. Let g : Y → X be a homotopy inverse of f so that g ◦ f ' 1X and f ◦ g ' 1Y . By the theorem
above (g ◦ f )∗ = g∗ ◦ f∗ : π1 (X, p) → π1 (Y, f (p)) → π1 (X, g ◦ f (p)) deviates from (1X )∗ (which is the
identity homomorphism π1 (X, p) → π1 (X, p)) by an isomorphism. Thus g∗ ◦ f∗ is an isomorphism and
g∗ has to be surjective and f∗ has to be injective. Thus, since g is a homotopy equivalence too, g∗ has to
be injective. It follows that g∗ is an isomorphism and consequently f∗ is also an isomorphism.
Covering spaces and calculations
In this section we will consider one technique for calculating the fundamental group of a space. It is
neither algorithmic nor applicable in all cases, but does lead to a calculation of the fundamental group
of the Klein bottle for example.
The story starts with the following observation:
Suppose that G is a discrete group acting on a simply connected space X. Fix a base point p ∈ X and
let q : X → X/G denote the quotient map from X to the orbit space. Next, for each g ∈ G choose a path
p → gp. This is possible since X is path-connected and since X is even simply connected any two choices
of paths p → gp are homotopic. We note that q ◦ α is a loop in X/G based at q(p), since q(gp) = q(p).
This give us a function
φ : G → π1 (X/G, q(p)), φ(g) = hq ◦ αg i = q∗ (hαg i).
This function is in fact a group homomorphism. Indeed, αgh is a path p → (gh)p, but so is the composite
p → gp → g(hp). Here gαh is the path t 7→ g(αh (t)). Since X is assumed to be simply connected the
two paths are homotopic, so hαgh i = hαg .gαh i. Also, note that q ◦ (gαh ) = q ◦ αh , since g(αh (t)) and
αh (t) are in the same orbit of the action.
This leads to
q∗ (hαgh i) = q∗ (hαg .gαh i) = q∗ (hαg i).q∗ (hgαh i) = q∗ (hαg i).q∗ (hαh i) = φ(g).φ(h).
Thus φ respects the composition rule. The other requirement of a group homomorphism is that it maps
the unit element 1 of G to the unit element hcq(p) i of π1 (X/G, q(p)). We check this:
φ(1) = q∗ (hα1 i) = q∗ (hcp i) = hcq(p) i.
We have used that both α1 and cp are paths p → 1p = p in X.
Covering actions
We cannot generally expect the homomorphism φ (of the previous section) to be injective, let alone an
isomorphism. Suppose for example that gp = p for some g 6= 1 in G, then φ(g) = hcq(p) i. A minimal
requirement for φ to be injective is that the action of G on X is free: gx = x for some x ∈ X only if
g = 1. If G is finite and X is Hausdorff, this turns out to be the only thing we need to assume to make
φ an isomorphism.
In general we have to assume that the action is a covering action (a term not used in Armstrong’s book):
Definition 3.1 A group action
` G × X → X is a covering action if any point in X has an open neighborhood U such that GU = g∈G gU .
Here are three general types of actions which are covering actions
1. If H is a discrete subgroup of a topological group G, then the action H × G → G, (h, g) 7→ gh
(multiplication in G) is a covering action.
To verify the condition let first p = 1 the unit element of G (and H). Since H is discrete there is
an open set O in G such that O ∩ H = {1}. Continuity of the map m : G × G → G, m(x, y) = xy −1
and the fact that (1, 1) ∈ m−1 (O) allows us to find a basic open set V × V ⊂ m−1 (O) containing
(1, 1). The image of V × V under m is V · V −1 ⊂ O and contains 1.
Suppose that x ∈ hV ∩ h0 V, where h, h0 ∈ H. Then x = hv = h0 v 0 , some v, v 0 ∈ V . This gives
h−1 h0 = v(v 0 )−1 ∈ H ∩ O = {1}, so h = h0 and hV and h0 V are thus disjoint unless h = h0 .
If p = x is a general point of G then V x is an open set containing x and hV x ∩ h0 V x = ∅, unless
h = h0 .
2. Suppose G is finite and acts freely on a Hausdorff space X, then the action is a covering action.
To verify this, fix p ∈ X and g ∈ G, g 6= 1. Then p 6= gp and since X is Hausdorff we can find
open disjoint sets Og containing p and Og containing gp. Then U g = Og ∩ g −1 Og is an open set
containing p such that U g and gU g are disjoint: the first is contained in Og the second one in Og
and these two are disjoint.
Next put U = g∈G U g and suppose gU ∩ g 0 U 6= ∅. Multiplying by g −1 then gives U ∩ g −1 g 0 U 6= ∅.
−1 0
But U ⊂ U g g and g −1 g 0 U ⊂ g −1 g 0 U g
g −1 g 0 = 1, i.e. g = g 0 .
−1 0
where U g
−1 0
and g −1 g 0 U g
−1 0
are disjoint unless
3. Suppose G acts freely on a metric space X through isometries, i.e. multiplication by g ∈ G
preserves distances: d(gx, gy) = d(x, y). If, in addition, each point p has a neighborhood O such
that O ∩ Gp = {p}, (that is p is open in Gp )then the action is a covering action.
To verify this choose > 0 such that B (p) ⊂ O. Then d(gp, p) > when g 6= 1 and U = B/2 (p)
has the required property. Indeed, gB/2 (p) = B/2 (gp) and multiplying gB/2 (p) ∩ hB/2 (p) by
g −1 gives the homeomorphic B/2 (p) ∩ B/2 (g −1 hp). But this set is empty if g −1 h 6= 1 since then
d(p, g −1 hp) > .
Here are three examples of the situations above:
1. Z is a discrete subgroup of R (which is a topological group with respect to ordinary addition). Here
the orbit space R/Z is homeomorphic to S 1 through the exponential map.
As R is simply connected we have the homomorphism φ : Z → π1 (S 1 , 1). To spell out the details
of this map we take p = 0, and we choose αn to be the linear path 0 → n defined by αn (t) = tn,
which then maps to the loop exp(tn2πi) based at 1 in S 1 , i.e n full revolutions on the circle.
2. Let C2 be the cyclic group of order 2 (written multiplicatively) with generator τ . A free action of
this finite group on the Hausdorff space S n is defined by τ · x = −x. Here the orbit space is the
projective space Pn of dimension n.
We will see that S n is simply connected if n ≥ 2. With this proviso, and taking p = en+1 , the
north-pole, we have the homomorphism φ : C2 → π1 (Pn , [en+1 ]).
To spell out the definition of φ we choose a path from en+1 to τ en+1 = −en+1 , for example
α(t) = cos(tπ)e1 + sin(tπ)en+1 . When this path is transported to Pn using the quotient map it
becomes a loop at [en+1 ].
3. Define A and B to be the isometries A(x, y) = (x, y + 1) and B(x, y) = (x + 1, 1 − y) of the metric
space R2 . They generate a subgroup of the group of all homeomorphisms of R2 , which we now
We notice that ABA(x, y) = AB(x, y + 1) = A(x + 1, −y) = (x + 1, 1 − y) = B(x, y), so that
ABA = B or BA = A−1 B. Hence we can move an A passed a B at the expense of inverting it. It
follows that An B m Ar B s = An+(−1) r B m+s . We next show that An B m = Ar B s only if n = r and
m = s by calculating
A (x + m, y)
(x + m, y + n)
if m is even
n m
A B (x, y) =
An (x + m, 1 − y)
(x + m, n + 1 − y) if m is odd.
Thus the group G generated by A and B is G = {An B m |n, m ∈ Z} and multiplication is performed
according to the rule An B m Ar B s = An+(−1) r B m+s and the action on R2 is as described above.
n m
p distance between (x, y) and A B (x, y), with (n, m) 6= (0, 0),
√ that the action is free! As the
is m + n ≥ 1 if m is even and m + (n + 1 − 2y) ≥ 1 if m is odd the condition in 3 above is
It’s not to hard to convince one self that all orbits of this action passes trough I 2 exactly once, with
the exceptions that (0, y) is in the same orbit as (1, 1 − y) and (x, 0) is in the same orbit as (x, 1).
The quotient space R2 /G is homeomorphic to the Klein-bottle K.
As R2 is simply connected we have a homomorphism φ : G → π1 (K, [0, 0]).
Covering projections
We now return to definition 3.1 and analyze its implications on the (open) quotient map q : X → X/G.
Let U be an open set containing x as in the definition of a covering action.`Then V = q(U ) is an open
neighborhood of Gx in X/G such that q −1 (V ) is a disjoint union, namely g∈G gU of open sets in X.
Furthermore, the restriction of q to each of these open sets is an open bijection q|gU → V = q(U ) and
hence a homeomorphism.
To focus on the relevant information in this situation, we formalize this as follows. Suppose p `
is a surjective map. An open subset V of Y is evenly covered by p if p−1 (V ) is a disjoint union j∈J Uj ,
such that the restriction of p to each of the Uj :s is a homeomorphism p| : Uj → V . It follows from this
that the fiber p−1 (y) is a discrete subspace of X consisting of one point in each of the disjoint open sets
Uj , j ∈ J.
We next define a surjective map p : X → Y to be a covering projection (X a covering space of Y, Y is the
base space of p) if each point of Y has an open neighborhood evenly covered by p. The prime example of
covering projection is the case when p is the quotient map X → X/G of a covering action of a group G
on X. (Actually, pretty much all covering projections arise in this way, but that’s another story.)
We now consider how covering projections relates to homotopy classes of paths in Y and X. We do this
by proving a sequence of lemmas.
A lift of a map f : Z → Y with respect to p : X → Y is a map f˜ : Z → X, such that f = p ◦ f˜. It’s also
called a factorization of f through p.
Lemma 3.1 Suppose Z is connected and p : X → Y is a covering projection. If f : Z → Y is a map
with image contained in an open set of Y evenly covered by p and x ∈ X and z ∈ Z are given points with
f (z) = p(x), then there is a unique lift f˜ of f with respect to p with f˜(z) = x.
Proof. By the assumptions we have f : Z → V where p−1 (V ) = j Uj , where each Uj is open in X.
Suppose x ∈ Uj0 . We have the homeomorphism q| : Uj0 → V . Observe that the connectivity of Z means
that f˜(Z) has to be contained in Uj0 , which actually determines f˜ as f˜ = (q|)−1 ◦ f (with f˜(z) = x). Lemma 3.2 Suppose p : X → Y is a covering projection and x is a fixed point of X. Then any map
F : I 2 → Y has a unique lift F̃ : I 2 → X with F̃ (0, 0) = x.
Proof. By assumption Y can be covered by open sets evenly covered by p. The inverse images of these
under F form an open cover of I 2 , which is a compact metric space. Consequently, by the Lebesgue’s
lemma, we can choose equidistant points on the x- and y-axises of I 2 so that each of the resulting small
squares in I 2 is mapped into an evenly covered open subset of Y by F .
Starting with the lower left hand square in this subdivision, there is, by the lemma above, a unique lifting
of F defined on this square with (0, 0) 7→ x. Moving to the next square on the right side of it we can
extend the lift uniquely to this square, again using the lemma, which further assures that the lifts agrees
on the line segment common for the two squares. Continuing in this fashion we can find a unique lift of
F on all of the squares of the bottom row in the subdivision of I 2 . We next move to the second row from
the bottom in the subdivision and find that we can repeat the argument to extend the partially defined
lift uniquely to include the second row,too. Repeating row by row we conclude that the unique lift F̃
with F̃ (0, 0) = x exists.
Lemma 3.3 Suppose p : X → Y is a covering projection and x is a fixed point of X. Then any map
α : I → Y has a unique lift α̃ : I → X with α̃(0) = x.
Proof. Similar to, but easier than, the proof of lemma 3.2.
(You can actually get lemma 3.3 from lemma 3.2 by considering the map I 2 → Y, (t, s) 7→ α(s). Details
are left to the reader.)
Covering projections and paths
Using the lemmas of the previous section we now draw the following conclusions on the relation between
paths in Y and paths in X in a covering projection:
Theorem 3.1 Suppose p : X → Y is a covering projection and x ∈ X a fixed point. Then
1. any path α in Y starting at p(x) in the image of a unique path α in X starting at x
2. if α ' β and β̃ is a lift of β starting at x, then α̃ ' β̃.
Proof. In view of lemma 3.3 only the second part requires an argument.
Let F : α ' β. Then F has, by lemma 3.2, a unique lift F̃ with F̃ (0, 0) = x.
To see that F̃ is a homotopy α̃ ' β̃ we observe first that p ◦ F̃ (0, s) = F (0, s) = p(x) and p ◦ F̃ (1, s) =
F (1, s) = α(1), so that F̃ (0, s) and F̃ (1, s) are lifts of constant paths. By uniqueness they are constant
too. Thus F̃ is a path homotopy between paths in X starting at x.
Now p ◦ F̃ (t, 0) = F (t, 0) = α(t) and p ◦ F̃ (t, 1) = F (t, 1) = β(t). Uniqueness implies that F̃ (t, 0) = α̃(t)
and F̃ (t, 1) = β̃(t).
The conclusion
We finally return to the starting point of the story:
Theorem 3.2 Suppose G acts on a simply connected space X by a covering action. Then, for any x ∈ X
the map
φ : G → π1 (X/G, q(x))
defined above is an isomorphism.
Proof. We first show surjectivity. Let hβi be in π1 (X/G, q(x)) represented by a loop q(x) → q(x). Let
β̃ be a lift of β starting at x. Then p ◦ β̃(1) = β(1) = p(x), so β̃(1) = gx, some g ∈ G. Since X is simply
connected αg ' β̃, since both are paths x → gx, and
= q∗ (hαg i) = q∗ (hβ̃i) = hβi.
Next, to show injectivity of a group homomorphism it suffices to show that only the unit element is
mapped to the unit element. So suppose φ(g) = hcq(x) i, i.e q ◦ αg ' cq(x) . Then αg is a lift of q ◦ αg while
cx is a lift of cq(x) . By 2 of Theorem 3.1 we have that αg ' cx , in particular gx = αg (1) = cx (1) = x. By
the condition on the action this implies that g = 1.
This theorem gives the following (non-algorithmic) method of trying to compute the fundamental group
of a space Y :
Find a simply connected space X and a covering action of a discrete group G on X such that
X/G is homeomorphic (or even less: homotopy equivalent) to Y .
The examples 1 – 3 above leads to the computations (once it’s know that S n is simply connected for
n ≥ 2)
1. π1 (S 1 ) ∼
= Z and a generator of π1 (S 1 , 1) is hexp(2πit)i.
2. π1 (Pn ) ∼
= C2 , when n ≥ 2 and a generator in π1 (P n ) is hq(cos(πt)e1 + sin(πt)en+1 )i.
3. π1 (K) ∼
= G, where G = {An B m | n, m ∈ Z} and An B m Ar B s = An+(−1) r B m+s . The elements
A and B corresponds to the classes of loops in K that you get from the paths a(t) = (t, 0) and
b(t) = (t, t) in I 2 by applying the quotient map from I 2 to K.
The fundamental group of a topological group
We have seen that the fundamental group of S 1 is abelian: the multiplication is commutative. This is no
coincidence since the fundamental group of (any path-component) of a topological group turns out to be
abelian. We show this.
First note that if 1 → 1 are loops in a topological group G, then there we could use the multiplication
in the group to get a second way of composing them. Define α ∗ β(t) = α(t) · β(t) where the dot denotes
multiplication in G. We will also show that hα ∗ βi = hαi.hβi.
α, β
Theorem 3.3 Suppose that G is a topological group and let 1 → 1 be loops at the unit element of G.
hαi.hβi = hα ∗ βi = hβi.hαi.
Thus the fundamental group of a topological group is abelian.
(1, 1),
(1, 0), (0, 1) →
Proof. Define a map F : I 2 → G by F (t, s) = α(t) · β(s). In I 2 let (0, 0) →
(0, 0) → (0, 1), (1, 0) → (1, 1) be the standard linear paths. Also let (0, 0) → (1, 1) be the linear diagonal
path. Applying F to these paths gives in turn, α, β, β, α and α ∗ β.
Since I 2 is a convex subset of R2 we have
1 .4 ' 5 ' 3 .2 ,
since all three are paths (0, 0) → (1, 1). Applying F to this gives
α.β ' α ∗ β ' β.α
Example. Since we have seen that the fundamental group of K is non-abelian it’s not possible to find
a continuous multiplication on it which makes it into a group.
Example Suppose G is a simply connected topological group (such as S 3 for example), then any normal
discrete subgroup H has to be abelian. In fact, normality of H ensures that G/H is a topological group,
so H ∼
= π1 (G/H) has to be abelian.
Example The subgroup Q8 = {±1, ±i, ±j, ±ij} of S 3 is not normal. (Try to prove this directly, it’s
not hard.)